m = mass of the partner which the cheerleader lifts = 59.6 kg
h = height to which the partner is lifted by the cheerleader = 0.749 m
g = acceleration due to gravity = 9.8 m/s²
work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.
W = work done by the cheerleader in lifting the partner
PE = potential energy gained
so W = PE
potential energy is given as
PE = mgh
hence
W = mgh
inserting the values in the above formula
W = 59.6 x 9.8 x 0.749
W = 437.5 J
this is the work done in lifting the partner once.
the cheerleader does this 30 times , hence the total work done is given as
W' = 30 W
W' = 30 x 437.5
W' = 13125 J
Answer:
Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.
Explanation:
<em>A</em><em>N</em><em> </em><em>E</em><em>L</em><em>E</em><em>C</em><em>T</em><em>R</em><em>I</em><em>C</em><em>I</em><em>T</em><em>Y</em><em> </em><em>M</em><em>O</em><em>T</em><em>O</em><em>R</em><em> </em><em>U</em><em>S</em><em>E</em><em>S</em><em> </em><em>E</em><em>L</em><em>E</em><em>C</em><em>T</em><em>R</em><em>I</em><em>C</em><em>I</em><em>T</em><em>Y</em><em> </em><em>T</em><em>O</em><em> </em><em>R</em><em>U</em><em>N</em><em> </em><em>O</em><em>N</em>
We could use the formula for the Power supply in order to find the wattage capacity and it would be:
P = V²/R or P = V * I
Hope this helps!
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.
First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055
Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3
Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.
T then is equal to 20.20 N
